A small ball of mass 1 kg and charge $q = \frac{2}{3} μ C$ is placed at the centre of a uniformly charged sphere of radius 1 m and charge $Q = \frac{1}{3} m C$ . A narrow smooth horizontal groove is made in the sphere from centre to surface as shown in figure. The sphere is made to rotate about its vertical diameter at a constant rate of $\frac{1}{2 π}$ revolutions per second. Find the speed w.r.t. ground (in $m/s$ ) with which the ball slides out from the groove. Neglect any magnetic force acting on ball.

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Detailed Solution

$\frac{1}{2} m v_{r}^{2} = (\frac{3}{2} \frac{K Q q}{R} - \frac{K Q q}{R}) + {\int^{}}_{0}^{R} m ω^{2} x d x = \frac{K Q q}{2 R} + \frac{m ω^{2} R^{2}}{2} S u b s t i t u t i n g v a l u e s, v_{r}^{2} = 2 + 1 = 3 v_{t} ω = r = 1 v_{n e t} = \sqrt{V_{r}^{2} + V_{t}^{2}} = \sqrt{3 + 1} = 2 m / sec$